Cho hàm số f(x) liên tục trên R và thỏa mãn \(xf({{x}^{3}})+f(1-{{x}^{2}})=-{{x}^{10}}+{{x}^{6}}-2x,\forall x\in \mathbb{R}\).

* Hướng dẫn giải

Ta có \(xf\left( {{x}^{3}} \right)+f\left( 1-{{x}^{2}} \right)=-{{x}^{10}}+{{x}^{6}}-2x,\,\forall x\in \mathbb{R}\,\,\,\,\left( 1 \right)\)

       \(\Leftrightarrow {{x}^{2}}f\left( {{x}^{3}} \right)+xf\left( 1-{{x}^{2}} \right)=-{{x}^{11}}+{{x}^{7}}-2{{x}^{2}}\)

      \(\Rightarrow \int\limits_{-1}^{0}{{{x}^{2}}f\left( {{x}^{3}} \right)dx+\int\limits_{-1}^{0}{xf\left( 1-{{x}^{2}} \right)}dx=\int\limits_{-1}^{0}{\left( -{{x}^{11}}+{{x}^{7}}-2{{x}^{2}} \right)}dx}=\frac{-17}{24}\)

Xét \({{I}_{1}}=\int\limits_{-1}^{0}{{{x}^{2}}}f\left( {{x}^{3}} \right)dx\) đặt \(u={{x}^{3}}\Rightarrow du=3{{x}^{2}}dx\Rightarrow \frac{1}{3}du={{x}^{2}}dx\)

Đổi cận: \(\left\{ \begin{array}{l}
x =  - 1 \Rightarrow u =  - 1\\
x = 0 \Rightarrow u = 0
\end{array} \right.\)

\(\Rightarrow {{I}_{1}}=\frac{1}{3}\int\limits_{-1}^{0}{f\left( u \right)}du=\frac{1}{3}\int\limits_{-1}^{0}{f\left( x \right)dx}\)

Xét \({{I}_{2}}=\int\limits_{-1}^{0}{xf\left( 1-{{x}^{2}} \right)}dx\) đặt \(u=1-{{x}^{2}}\Rightarrow du=-2xdx\Rightarrow \frac{-1}{2}du=xdx\)

Đổi cận: \(\left\{ \begin{array}{l}
x =  - 1 \Rightarrow u = 0\\
x = 0 \Rightarrow u = 1
\end{array} \right.\)

\(\Rightarrow {{I}_{2}}=-\frac{1}{2}\int\limits_{0}^{1}{f\left( u \right)}du=-\frac{1}{2}\int\limits_{0}^{1}{f\left( x \right)}dx\)

\(\Rightarrow \frac{1}{3}\int\limits_{-1}^{0}{f\left( x \right)dx}-\frac{1}{2}\int\limits_{0}^{1}{f\left( x \right)}dx=\frac{-17}{24}\,\,\left( 2 \right)\)

Trong (1) thay x bởi -x ta được: \(-xf\left( -{{x}^{3}} \right)+f\left( 1-{{x}^{2}} \right)=-{{x}^{10}}+{{x}^{6}}+2x,\,\,\,\left( 3 \right)\)

Lấy (1) trừ (3) ta được: \(xf\left( {{x}^{3}} \right)+xf\left( -{{x}^{3}} \right)=-4x\)

                                      \(\Rightarrow {{x}^{2}}f\left( {{x}^{3}} \right)+{{x}^{2}}f\left( -{{x}^{3}} \right)=-4{{x}^{2}}\)

                                  \(\Rightarrow \int\limits_{-1}^{0}{{{x}^{2}}f\left( {{x}^{3}} \right)dx+\int\limits_{-1}^{0}{{{x}^{2}}f\left( -{{x}^{3}} \right)}dx=\int\limits_{-1}^{0}{-4{{x}^{2}}}dx}=\frac{-4}{3}\)

                                \(\Rightarrow \frac{1}{3}\int\limits_{-1}^{0}{f\left( x \right)dx}+\frac{1}{3}\int\limits_{0}^{1}{f\left( x \right)}dx=\frac{-4}{3}\,\,\left( 4 \right)\)

Từ (2) và (4) suy ra \(\int\limits_{-1}^{0}{f\left( x \right)}dx=\frac{-13}{4}\)